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The special theory of relativity describes the motion and dynamics of objects moving at significant fractions of the speed of light.

The text says "By symmetry RU = SV and so these events are equidistant according to B. However, the signal RQ was sent before the signal SP and so B concludes that the event Q took place well befor …
answered Mar 24 '17 by robphy
From your setup, there is symmetry to exploit. If you break up the T-to-3T trip in half, you can see [by drawing a spacetime diagram] that you have four congruent legs to your trip. You just calculate …
I assume that these questions deal specifically with the [simple] twin paradox, where the twins reunite after separation with one twin inertial and the other non-inertial [but piecewise-inertial]. B …
answered Apr 14 '17 by robphy
Radar methods are not the only way to establish position and time coordinates. One can use Einstein's rods and clocks. For inertial motion in Special Relativity, they lead to identical coordinate assi …
answered Mar 24 '17 by robphy
The short answer: Length Contraction involves the apparent spatial separation of two parallel [timelike] worldlines, marking the ends of a stick. The contraction depends only on the relative-speed …
Here's a spacetime diagram drawn on rotated graph paper (so that one can more easily measure displacements in time and space along segments and so that one can visualize the orthogonality between an o …
While $\lambda= \frac{ c-v}{c}\gamma\lambda_o$ is correct [where $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$], $\lambda(laser)= \frac{ c-v}{c}\gamma^2\lambda_o$ is incorrect since $\lambda(laser)\neq \gamma … answered Apr 17 '17 by robphy Here's a spacetime diagram on rotated graph paper which may help visualize the result you obtained and help develop a strategy for getting the result from time-dilation and length contraction. The tr … answered Mar 13 '18 by robphy Here is a spacetime diagram [drawn on rotated graph paper] of the back of the stick showing two reference frames, one for the back of the stick (through the origin event), and the other for the fro … answered Feb 21 by robphy Here are two geometrical ways to calculate the invariant mass in this totally-inelastic collision. I will use the example cited by Rumplestillskin: A ball of mass 16kg with velocity 3/5c inelasti … answered Mar 15 '17 by robphy Consider the worldlines of the electrons (in motion, say with$v=0.25c\$ for convenience) and the protons (at rest) in Alice's frame where the wire is neutral. Let's draw this on a spacetime diagram on …